MECH 430 – FLUIDS 2
Flow of a compressible fluid in a constant area duct with friction
For an incompressible fluid where the density remains constant, friction results in a pressure drop in the flow direction. For a compressible flow, flow, viscous dissipation heats up the gas resultin resulting g in a density change. Thus, the the flow flow velocity velocity and pressure pressure can increase or decrease depending on the rate of expansion due to density decrease and the convective mass flux through through the cross section of the pipe. As in in the case of area change, friction has opposite effect if the flow is subsonic or supersonic. Let us first review the steady flow of an incompressible fluid in constant area pipe with friction (which results in a wall shear stress ). Referring to the sketch below: w
L
p + dp
p
w The conservation of mass for steady flow gives constant, the u u=constant and since flow velocity is constant throughout the the length of the duct. The momentum equation gives: pA p dp A w Aw 0 where A is the area of the duct ( A
D2
for a circular duct) and Aw is the wetted area 4 ( Aw DL for a circular pipe). Rearranging the momentum equation gives:
dp w
Aw A
4
w L
D
where we have assumed a circular pipe for for convenience. The wall shear stress stress is given by:
u r r
R
u r r
R
w
where is the coefficient of viscosity and
is the velocity gradient at the wall (if
we assume a laminar flow). The velocity profile for for steady laminar incompressible incompressible flow in a circular pipe is given by:
r 2 2 1 u R u
where u is the mean flow flow velocity (i.e. (i.e. the classical classical Hagen-Poiseuille Flow.) This parabolic velocity profile gives the wall shear stress as:
u r r
w
4 u
R
R
8 u D
The negative sign denotes the direction of w that gives the parabolic profile and we have w when we write the already considered the appropriate sign (i.e. direction) of momentum equation. Thus, we write:
dp L where Re D
p1 p2 L
4 w
D
32 u
D 2
64 u 2 1
Re D 2 D
u D
is the the Reynolds number based on the pipe diameter. In general, we define a friction factor “ f ” as:
f
4 w
u2 2
where f Re D is determined from experiments. (Moody diagram where f is is given for a wide range of Reynolds number and wall roughness). For the present case of laminar Poiseuille Flow:
2
64
f
Re D
We may write the pressure drop as: p1 p 2 L
u 2 1 f u 2 Re D 2 D D 2 64
For incompressible flow, we need not worry about the energy equation since constant and the problem is a dynamic problem with pressure forces balancing the friction forces.
Compressible Flow (Fanno Flow) For steady compressible flow in a constant area, adiabatic duct with friction as illustrated below, i.e.:
u p T
P+dp d T+dT u+du
w dx
The conservation of mass gives:
uA constant m d
du u
0
since dA=0 for a constant area duct.
The conservation of momentum can be written as:
3
1.
mdu dpA w Dsx
2.
where S and A are the perimeter and the area of the cross-section of the tube, respectively. Defining a friction factor:
f
w u2 2
and a hydraulic diameter “DH” by:
D H
4 * area wetted perimeter
4A
S
Equation 2 becomes:
udu dp
u 2 4 fdx 2
D H
3.
If the flow is adiabatic, the energy equation is given by:
h
u2 2
ho constant
Note that there is no work done by the viscous stress w since the velocity at the wall vanishes. The no slip (i.e. u=0) condition at the wall generates a velocity profile resulting in viscous dissipation. The viscous heating is at the expense of the kinetic energy decrease in the flow but since there is no heat transfer, the viscous heating remains in the flow. In the present assumption of a quasi one dimensional flow, the flow velocity is uniform across the cross section of the duct. We should consider the flow velocity as an averaged value to permit the presence of a velocity gradient for the viscous stress to occur. In an inviscid flow, there will be no shear stress at the wall. For adiabatic (not isentropic) flow where h
u2 2
H 0 cons tan t
we get
dh+udu=0
4
Since h=c pT, and c p
R 1
and c 2 RT , the energy equation can be written as:
dT
1 M 2
du
T The equation of state for a perfect gas is:
u
0
4.
p RT and hence
dp p
d
dT
5.
T
Rewriting Equation 3 as:
udu p and noting that c 2
p
dp
p
u 2 4 fdx 2 p D H
, we obtain an expression for
dp p
M
2
du u
dp p
as
M 2 4 fdx
From Equations 1, 4, 5 and 6, we can get an expression for
du u
M 2
6.
2 D H
4 fdx
du u
as
7.
21 M 2 D H
From Equation 1, we see that:
M 2 4 fdx u 21 M 2 D H
d
du
From Equation 4, we get
dT T
1 M 2
5
du u
8.
and substituting Equation 7 into the above yields
dT T
- -1 M 4 4 fdx 2 1- M 2
9.
D H
Substituting Equations 8 and 9 into Equation 5, gives:
dp
p
M 2 1 1 M 2 4 fdx 21 M 2
D H
To get an expression for the Mach number, we note M
dM
M and since c 2 RT ,
dc c
du u
u c
10.
, thus
dc c
1 dT 2 T
Hence,
dM M
du
u
1 dT 2 T
And substituting Equations 7 and 9 into the above, gives:
dM M
M 2 1
1 2
4 fdx
M 2
21 M 2
D H
11.
For the variation of the stagnation pressure along the pipe, we note that the definition of the stagnation pressure is the pressure one would obtain if we decelerate a flow isentropically to zero velocity. Thus,
1 2 1 M p 2
po
1
and differentiating, we get:
dpo po
dp p
1
6
M 2 1 2
dM M 2
M
The variation of p and M along the pipe is given by Equations 10 and 11, thus, we get:
dpo
po
M 2 4 fd x
12.
Dh
2
The variation of entropy along the pipe can be found from:
Tds = dh-vdp
with h c p T
RT 1
, pv=RT , the above becomes:
ds c p
Equations 9 and 10 give the variation of
dT
T
dT
1 dp p
and
dp
T p entropy along the pipe the following expression:
, thus we obtain for the change in
1 M 2 4 fd x c p 2 DH ds
ds
or
R
M 2 4 f 2 D H
dx
dpo
13.
po
c p 1
R . As in the case of the normal shock, the entropy increase can be correlated to the stagnation pressure loss due to viscous dissipation. since
Equations 7-13 give the variation of u, , T, P, M, po and s along the pipe dpo ds 2 respectively. Except for and , all the equations contain the term (1-M ) in the po R 2
denominator. Thus, depending on whether the flow is subsonic M<1, and (1-M )>0 or 2 supersonic, M>1 and (1-M ) <0, the various thermodynamic states can either increase or
7
decrease due to friction. The table below summarizes the effect of friction on the fluid and thermodynamic states. The sign + or – denotes whether the variable increases or decreases along the pipe (i.e. increased value of x or positive dx).
Variable
Subsonic
Supersonic
+ + -
+ + +
+
+
u M p T Stagnation pressure po Entropy s Velocity Mach Number Pressure Temperature Density
The difference between subsonic and supersonic flow is a result of the competition between the rate of expansion due to viscous heating of the flow and the rate of mass convected through the cross section of the pipe. Thus, for subsonic flow, the rate of expansion (characterized by the sound speed) dominates, and the density decreases resulting in an increase in the flow velocity, a decrease in temperature and pressure due to the expansion. The reverse happens for supersonic flow when the rate of convection (characterized by the flow velocity) dominates over the rate of expansion. However, for the stagnation pressure and entropy, both subsonic and supersonic flows indicate the same behavior of loss in stagnation pressure and increase in entropy from irreversible viscous dissipation.
Integration of the equations To get the variation of the various state variables along the duct, the differential equations have to be integrated. The important one to start is the variation of the Mach number with distance given by Equation 11. Rewriting it in the following form: 4 fdx
D H
1 M dM 2
M 4 1
2
1
2
2
M
we can integrate the above using the method of partial fraction. The above can be expanded as:
8
4 fdx
D H
2
dM M 4
1 dM
2
2
2
M
1
1
2
2
dM
2 1
2 1 M 2
Since friction always drives the flow towards M =1, we integrate the above from 2 * 0< x
1
1 2 2 * 2 M 4 fL 1 M 2 ln 1 2 M 2 D H 1 2 M
14.
The above equation says that if we have a duct where at x=0, the Mach number is * M, then at x=L , M=1 and the flow is choked. Equation 14 is computed for a range of Mach numbers and tabulated (Fanno flow tables). So, if we have a pipe of length L and inlet Mach number M1, and if we want to know the Mach number at L, we do the following:
M 1
M
*
L
L
LM1
Referring to the sketch above:
L L* M1 L* M 2 or
9
M2
4 fL
D H
4 fL* M 1
D H
4 fL* M 2
D H *
* 1
We know “L” and “M1”, we can look up the table for L such that
4 fL M 1
D H
corresponds
to M 1 . Then, we can find the value for: 4 fL* M 2
D H
4 fL* M 2
D H
4 fL
DH
We can then find the value of M2 corresponding to the value of *
4 fL* M 2
D H
.
Equation 14 gives “M” as a function of L . If we want to know the other flow variables dp dT like p, T, , etc. we can integrate the other equations for , , etc. However, it is p T more convenient to express them in terms of the Mach number so that once “M” at a section is known, we can find the other variables. From Equations 10 and 11, we write:
1 1 M 2 M 2 4 fdx D H 21 M 2
dp
p dM M 2 1 2 1 M M 4 fdx 2 2 D H 1 M 2 and simplifying the above yields:
1 1 M 2 dM 2 p 1 2 M 2 2 1 M 2
dp
Using partial fraction, the above can be written as:
1
dp p
2
dM 2
2 M
2 1
2 1
10
dM 2
2
M 2
which integrates to yield:
p p
*
1
2 1
M
1 1 2
15.
M 2
*
The above equation gives the pressure “ p” at “M’ and p = p when M=1. Similarly, we can find expressions for the other variables in terms of the Mach number as follows:
2
1 c T * c * 1 2 M 2 1 2 T
u u*
*
M
u
1
17.
1 2 M 2 1 2
*
u
16.
1
M
2 1
1 2
M 2
1
18.
19.
1
1 2 2 M 2 1 po 1 2 po * M 1
1
* 1 s s 2 ln M 1 c p 2 2 M M 2 1 2
11
1
2
20.
Equations 15 to 20 are also computed for various Mach numbers and tabulated together 4 fL* with in the Fanno tables for convenience in solving problems of compressible flow D H in ducts with friction.
The Fanno Line The adiabatic flow of a compressible fluid in a constant area duct with friction satisfies the conservation of mass and energy i.e.
constant
u
d
h
u2 2
du u
0
ho constant
dh udu 0 From these two equations, we can obtain an expression for the variation of the enthalpy (or equivalently the temperature) and the entropy along the duct. From the Tds equation, i.e.:
Tds
dh vdp
we get: ds c p
dh h
1 dp
p
and from the equation of state p RT , we get:
dp p
d
dT T
Hence,
12
d
dh h
ds c p
dh
h
1 d
dh 1 dh 1 d h h
From the conservation of mass, we write: d
du u
du 2
2u 2
From the energy equation, we note that:
u2 2 and
du 2 2
h o h
dh d ho h since ho constant.
Thus, we may write: d
h 2ho h
d ho
and the equation for the entropy becomes: ds c p
1 dh h
h ho h
1 d ho
2
We can integrate the above as following: s
ds
s*
c p
1
h
dh
h*
h
1 2
ho h
d ho h
ho h*
ho h
and obtain:
s s * c p
Since ho
1
1
h
*
ln
h
1 2
ho h * ho H
ln
h* , we can eliminate ho in the above equation and obtain the equation for
2 the Fanno Line as:
13
1 1 1 2 2 s s h h 2 1 ln * * c p h h 1 2
*
21.
The Fanno line gives the variation of the enthalpy as a function of the entropy for * a given mass flow rate in and stagnation enthalpy ho (or equivalently h ). The lower branch is for supersonic flow and the upper branch is for subsonic flow. For any given value of M, friction will drive the flow towards sonic conditions M=1 where the entropy is a maximum. To prove that entropy is a maximum when the flow becomes sonic, one ds could differentiate Equation 20 and equate 0 and solve for M. Alternately, we do dM the following: from the Tds equation, we write:
Tds dh
1
dp
and from the energy equation dh udu, we get:
Tds udu From the conservation of mass u
1
dp
constant , we get : du ud 0
If we express s, p , we can write: 1 ds dp ds 2 dp c s p s p p s
d s, p
since the sound speed “c” is defined by:
c2
p s
Replacing d in the continuity equation by the above expression yields : du ud du
u c
2
ds 0 s p
dp u
14
22.
From the Tds equation, we get:
dp udu Tds and replacing dp in Equation 22 by the above and rearranging gives:
T c 2 s p 0
du 1 M 2 uds
Since
T is always positive and c2
is s p
23.
always negative (since density decreases
when heat is added at constant pressure and heat added means entropy increases, i.e. 0 as ds>0 at constant “ p”). The bracketed term is always positive, i.e.:
0 c s p
pT 2
Thus, as M 1 , ds must vanish, thus “ s” corresponds to an extremum (maximum) value when M=1. Another alternate expression for the Fanno Line that involves the Mach number can be obtained as follows: From the Tds equation, we write:
ds
dh Rdp T p
and rearranging, we write:
dh ds
1
1 R dp
T p dh From the energy equation, we get: dh du
2
2
0
And the conservation of mass gives:
d
du u
15
1 du 2 2 u2
we can express dh as:
dh du
2
2
d
u2
And using the equation of state p RT , we can express:
d dp p
dT T
Hence,
dh u 2
And solving for the slope
dh ds
dT T
p 1 1M 2 2 u
as:
dh ds
or
dp , we get: dh dp dh
We can now write
dp d u 2 p
dh ds
1
R p p 2 1M 2 2 T p u u 1
M 2T
M 2 1
The above is an alternate form of the equation for the Fanno Line and it is clear that for dh subsonic flow where M<1, 0 which corresponds to the upper branch of the Fanno ds dh Line. For M>1, 0 , the locus of states is represented by the lower branch of the ds dh Fanno Line. When M=1, and s max . ds To plot a Fanno Line, we first note that the Fanno Line involves the conservation of mass and energy, i.e.
16
constant u m u2 h 2
ho constant
” and “ho ”. Combining conservation of Thus, a Fanno Line corresponds to a value for “m energy and mass, we get:
h
Where v
2v 2 m 2
ho constant
24.
1
and ho , we pick different values for is the specific value. So, given m and compute h from Equation 24. And Equation 21 gives “ s” as a function of “h”. Note 2 ho , thus the value of h* can be found for a given stagnation enthalpy “ho”. that h* 1
Effect of duct length and back pressure
The flow inside a duct with friction is influenced by the duct length and the back pressure of the exit of the duct. Consider first the case of subsonic flow in the duct. We assume the inlet to the duct is from a converging nozzle connected to a large plenum chamber where the stagnation pressure and temperature is “ po” and “T o”. Consider first just the nozzle. For a subsonic jet, the exit pressure must be the same as the back pressure. Thus, for a given exit pressure, the Mach number at the jet exit is given by the isentropic relationship 1
M 2 1 p 2
po
1
If we now connect a pipe to the exit of the converging nozzle, then the exit of the nozzle corresponds to the entrance to the duct. Let the end of the duct be connected to a large plenum chamber where the pressure can be controlled. The system is shown schematically in the sketch below.
17
stagnation conditions
Pe
plenum chamber PB
Po, To
Pe
a
Pe = PB
b
Pe = PB
c
Pe = P* = P B Pe = P* PB < P* ho
Me
Me Me = 1
a b
c
decreasing m
For curve “a”, the flow is subsonic throughout the duct, the exit pressure pe=p B . The 4 fL 4 fL* value of is less than the value corresponding to the inlet Mach number M 1. D H D H When we lower the back pressure p B, the flow is similar but the inlet Mach number to the duct (hence the mass flow rate) is increased. When the back pressure is reduced to a * value where the flow is choked at the exit, then pe=p =p B and M e=1. For this case, the 18
*
duct length corresponds to the critical value L for the inlet Mach number (curve “c”). If * the back pressure is further decreased, i.e. p B
since the flow is not choked at the exit and pe
B . The under expanded jet will adjust to ambient pressure via expansion waves. When the back pressure is increased to the exit pressure, i.e. pe= p B, then the jet exit pressure matches the ambient pressure and the jet is parallel. When p B>pe , but not significantly greater, then conditions inside the duct remains unchanged and adjustment to ambient pressure occurs downstream of the jet exit via oblique shock waves. When p B increased further, the oblique shock becomes stronger by increasing its angle to the flow. The limiting condition will be when a normal shock occurs just at the duct exit so that the pressure downstream of the normal shock matches the back pressure p B. When p B increases beyond this limiting value, the shock moves into the duct and becomes stronger since the Mach number is higher. Downstream of the shock is subsonic, thus the exit pressure will have to be equaled to the ambient pressure. The location of the shock is such that this condition is satisfied. The above discussion is for the case when the duct length is less than the critical value corresponding to the inlet Mach number. * If the duct length is the critical value, then pe= p and M e=1. Again, when * pe= p >p B , the under expanded sonic jet will undergo further expansion downstream of the * duct exit to adjust to the ambient pressure. If p B= p , the same kind of events as described * earlier for the case of LL for the inlet Mach number, a normal shock will be formed inside the duct. The position of the shock is such that the exit pressure matches the ambient pressure. If the duct is very long, then the maximum Mach number the subsonic flow downstream of the shock inside the duct can accelerate to is M=1. * * Thus, the flow is chocked at the exit, pe= p . If p B= p , further adjustment occurs 19
*
downstream of the duct exit. For pe= p , there is only one location where the shock inside the duct is located. For supersonic inlet flow and a very high p B so that a shock is driven inside the duct, the process is illustrated in the h-s diagram below.
ho
M< 1
isentropic flow in nozzle
20
Me , Pe = PB
normal shock transition M> 1
